>On Aug 31, 1:06 am, Steven D'Aprano <steve>+comp.lang.pyt...@pearwood.info> wrote:>> The population variance is given by:>>>> ?^2 = ?(x - µ)^2 / n>>>> with µ = population mean, the summation being over all the x in the>> population.>>>> (for brevity, I haven't attempted to show subscript-i on the x).>>>> If you don't have the entire population, you can estimate the variance with>> the sample variance:>>>> s^2 = ?(x - m)^2 / n (Eq. 1)>>>> where m = sample mean (usually written as x bar), and the sum is over all>> the x in the sample. A second estimator is:>>>> s^2 = ?(x - m)^2 / (n-1) (Eq. 2)>>>> which some people prefer because it is unbiased (that is, the average of all>> the possible sample variances equals the true population variance if you>> use the (n-1) version).>>>> See alsohttp://mathworld.wolfram.com/SampleVariance.html>>>> I have a set of data with an (allegedly) known population mean µ but an>> unknown ?^2. I wish to estimate ?^2. Under what circumstances should I>> prefer Eq.1 over Eq.2?>>>> Or should I ignore the sample mean altogether, and plug the known population>> mean into one of the two equations? I.e.:>>>> s^2 = ?(x - µ)^2 / n (Eq. 3)>> s^2 = ?(x - µ)^2 / (n-1) (Eq. 4)>>>> Under what circumstances should I prefer each of these four estimators of>> ?^2 and what are the pros and cons of each?>>>> Thanks in advance,>>>> -->> Steven>>If you know mu, the population mean, divide by n.

That's a pretty complete answer.

But if you are cynical about the pop. mean being the pop. mean,then I suppose you could offer two answers, one using the observed mean. If you want tests using ANOVA, you use anunbiased estimate of the variance.

- To a question posted later: No, I don't see any place or justification for using the population mean in a computationwith (n-1) as denominator.